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I will take the contrary viewpoint. It matters what mathematics we consider to be beautiful and intuitive, because better mathematical intuition will make you b

by lacker 7y ago

I will take the contrary viewpoint. It matters what mathematics we consider to be beautiful and intuitive, because better mathematical intuition will make you better at solving mathematical problems. On that note, I think e^i pi = -1 is drastically overrated. That is typically the definition of exponentiating to a complex power, that e^ix = sin x + i cos x. Of course something can be beautiful if you take it to be an axiom. It’s like saying x=x is beautiful. Also it’s really just sticking -1 in the more-intuitive, more-useful formula so that it looks cool. A really beautiful formula is that x^p - x is divisible by p when x is an integer and p is prime....

kd5bjo 7y ago
Though it’s probably the most commonly used definition for complex exponentiation, e^i pi=-1 is not an axiom— it is derived from the basic rules of integer arithmetic expanded in scope until they make a closed system. And from this, we can derive all of the trigonometric functions using only algebra and no geometry. (See http://www.feynmanlectures.caltech.edu/I_22.html http://www.feynmanlectures.caltech.edu/I_22.html , particularly section 22-5)
thaumasiotes 7y ago
> I think e^i pi = -1 is drastically overrated. That is typically the definition of exponentiating to a complex power, that e^ix = sin x + i cos x. Yes, but it's the definition of exponentiating to a complex power because it is easily proven by reference to the real Taylor series of those functions. The complex definition is motivated by the real result, not -- as you imply -- the other way around. (Also, you have it backwards; e^ix is cos x + i sin x.)
username90 7y ago
You can easily derive e^ix = i cos x + sin x. Just look at the Taylor series for e^x and compare with the Taylor series for sin x + cos x. You'd see that they are basically the same, just that the trigonometric functions alternate their signs every 2 steps. How do we alternate the signs in e^x? We just add a factor of i, that way we get alternate signs every two steps but get an additional i on every odd step. Odd steps are sin, so just add i to sin and we are done: e^ix = i sin x + cos x. Now we also see that you remembered the wrong formula, it the sine part is imaginary since it corresponds to the odd factors.
- plopilop 7y ago
  (it's the other way round, e^(ix) = cos(x) + i sin(x))
  - Koshkin 7y ago
    Addition is commutative.
    - bzbarsky 7y ago
      Not sure whether that was meant as a joke or not, but the difference is which term is real and which is imaginary, not the order of the terms...
      - deleted 7y ago
        [deleted]
- contravariant 7y ago
  I have no problems with this route to show that e^ix = cos(x) + i sin(x) but to do it properly does require showing that this is the unique analytic continuation of the exponential function. The alternative method is to note that exp() is defined as the unique function satisfying exp(0) = 1 and dexp(ax)/dx = a exp(ax). So exp(ix) describes a point moving orthogonal to it's current position, which means it's a circle, and since it's speed is proportional to it's distance from the origin it is moving at a constant speed. So e^ix is just a point x radions around the unit circle.

by lacker 7y ago

kd5bjo 7y ago

Though it’s probably the most commonly used definition for complex exponentiation, e^i pi=-1 is not an axiom— it is derived from the basic rules of integer arithmetic expanded in scope until they make a closed system. And from this, we can derive all of the trigonometric functions using only algebra and no geometry. (See http://www.feynmanlectures.caltech.edu/I_22.html http://www.feynmanlectures.caltech.edu/I_22.html , particularly section 22-5)

thaumasiotes 7y ago

> I think e^i pi = -1 is drastically overrated. That is typically the definition of exponentiating to a complex power, that e^ix = sin x + i cos x. Yes, but it's the definition of exponentiating to a complex power because it is easily proven by reference to the real Taylor series of those functions. The complex definition is motivated by the real result, not -- as you imply -- the other way around. (Also, you have it backwards; e^ix is cos x + i sin x.)

username90 7y ago

You can easily derive e^ix = i cos x + sin x. Just look at the Taylor series for e^x and compare with the Taylor series for sin x + cos x. You'd see that they are basically the same, just that the trigonometric functions alternate their signs every 2 steps. How do we alternate the signs in e^x? We just add a factor of i, that way we get alternate signs every two steps but get an additional i on every odd step. Odd steps are sin, so just add i to sin and we are done: e^ix = i sin x + cos x. Now we also see that you remembered the wrong formula, it the sine part is imaginary since it corresponds to the odd factors.

plopilop 7y ago
(it's the other way round, e^(ix) = cos(x) + i sin(x))
- Koshkin 7y ago
  Addition is commutative.
  - bzbarsky 7y ago
    Not sure whether that was meant as a joke or not, but the difference is which term is real and which is imaginary, not the order of the terms...
    - deleted 7y ago
      [deleted]
contravariant 7y ago
I have no problems with this route to show that e^ix = cos(x) + i sin(x) but to do it properly does require showing that this is the unique analytic continuation of the exponential function. The alternative method is to note that exp() is defined as the unique function satisfying exp(0) = 1 and dexp(ax)/dx = a exp(ax). So exp(ix) describes a point moving orthogonal to it's current position, which means it's a circle, and since it's speed is proportional to it's distance from the origin it is moving at a constant speed. So e^ix is just a point x radions around the unit circle.